We can follow the proof of Theorem 1. It suffices to prove that
\(e^{-sL_0} e^{-sV}\)  is compact, and so, that \(e^{-sL_0} \chi _{\Omega _m}\)  is Hilbert–Schmidt.
That \(e^{-sL_0}\)  maps \(L^2\)  to \(L^\infty \)  implies, by the Dunford–Pettis theorem (see
[1]}), that there is, for each \(x\in X\) , a function \(f_x(\,\cdot \,)\in L^2 (X,d\mu )\) 
with
\( (e^{-sL_0} g)(x) =\langle f_x,g \rangle \) 
and
\( \sup _x \, \Vert f_x\Vert _{L^2} =\Vert e^{-sL_0}\Vert _{L^2\rightarrow L^\infty } \equiv C <\infty \) 
Thus, \(e^{-sL_0}\)  has an integral kernel \(K(x,y)\)  with
\( \sup _x \int \vert K(x,y)\vert ^2\, d\mu (y) = C<\infty \) 
(for \(K(x,y)=f_x(y)\) ). But \(e^{-sL_0}\)  is selfadjoint, so its kernel is complex symmetric, so
\( \sup _y \int \vert K(x,y)\vert ^2\, d\mu (x) =C <\infty \) 
Thus,
\( \int \vert K(x,y) \chi _{\Omega _m}(y)\vert ^2 \, d\mu (x) d\mu (y)\le C\mu (\Omega _m) <\infty \) 
and \(e^{-sL_0}\chi _{\Omega _m}\)  is Hilbert–Schmidt.
